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Is there a non-projective flat module over a local ring? Here I assume the ring is commutative with unit.

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$\mathbb{Q}$ is flat over $\mathbb{Z}_p$, but not projective.

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    $\begingroup$ Well done! So is $k((x))$ over $k[[x]]$... $\endgroup$
    – Bugs Bunny
    Commented Jul 21, 2010 at 21:39
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    $\begingroup$ So is the $\mathfrak{m}$-adic completion of any non-Artinian local ring. $\endgroup$
    – Graham Leuschke
    Commented Jul 21, 2010 at 22:25
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It is related to Bass' theorem. Flat modules are projective iff the ring is perfect. $p$-adic integers or formal power series are examples of local rings which are not perfect and have nonprojective flat modules.

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    $\begingroup$ But we can also make a slightly positive statement. Every local ring is semiperfect, and every finitely generated flat module over a semiperfect ring is projective. Moreover, since projective modules over a local ring are free, we see that every finitely generated flat module over a local ring is free. $\endgroup$
    – Manny Reyes
    Commented Jul 21, 2010 at 21:54
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    $\begingroup$ For those, who -- like me -- didn't know what a perfect ring was, see: en.wikipedia.org/wiki/Perfect_ring. $\endgroup$
    – Pete L. Clark
    Commented Jul 22, 2010 at 3:26

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